Chemical reaction: Ba(NO₃)₂ + H₂SO₄ → BaSO₄ + 2HNO₃.
V(H₂SO₄) = 250 mL ÷ 1000 mL/L = 0,25 L.
m(BaSO₄) = 0,55 g.
n(BaSO₄) = m(BaSO₄) ÷ M(BaSO₄).
n(BaSO₄) = 0,55 g ÷ 233,38 g/mol.
n(BaSO₄) = 0,00235 mol.
From chemical reaction: n(BaSO₄) : n(Ba(NO₃)₂) = 1 : 1.
n(Ba(NO₃)₂) = 0,00235 mol.
c(Ba(NO₃)₂) = n(Ba(NO₃)₂) ÷ V.
c(Ba(NO₃)₂) = 0,00235 mol ÷ 0,25 L.
c(Ba(NO₃)₂) = 0,0095 mol/L.
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I believe the answer would be A. equal temperature
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Given:
M = 0.0150 mol/L HF solution
T = 26°C = 299.15 K
π = 0.449 atm
Required:
percent ionization
Solution:
First, we get the van't Hoff factor using this equation:
π = i MRT
0.449 atm = i (0.0150 mol/L) (0.08206 L atm / mol K) (299.15 K)
i = 1.219367
Next, calculate the concentration of the ions and the acid.
We let x = [H+] = [F-]
[HF] = 0.0150 - x
Adding all the concentration and equating to iM
x +x + 0.0150 - x = <span>1.219367 (0.0150)
x = 3.2905 x 10^-3
percent dissociation = (x/M) (100) = (3.2905 x 10-3/0.0150) (100) = 21.94%
Also,
percent dissociation = (i -1) (100) = (</span><span>1.219367 * 1) (100) = 21.94%</span>
Answer: carbon dioxide detector
Answer:
b) The dehydrated sample absorbed moisture after heating
Explanation:
a) Strong initial heating caused some of the hydrate sample to splatter out.
This will result in a higher percent of water than the real one, because you assume in the calculation that the splattered sample was only water (which in not true).
b) The dehydrated sample absorbed moisture after heating.
Usually inorganic salts may absorbed moisture from the atmosphere so this will explain the 13% difference between calculated water percent the real content of water in the hydrate.
c) The amount of the hydrate sample used was too small.
It will create some errors but they do not create a difference of 13% difference as stated in the problem.
d) The crucible was not heated to constant mass before use.
Here the error is small.
e) Excess heating caused the dehydrated sample to decompose.
Usually the inorganic compounds are stable in the temperature range of this kind of experiments. If you have an organic compound which retain water molecules you may decompose the sample forming volatile compounds which will leave crucible so the error will be quite high.
